Mixing
Is the torus with one hole homeomorphic to the torus with two holes?
Clash Royale CLAN TAG#URR8PPP
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I would like to understand why the torus with one hole is not homeomorphic to the torus with two holes.
I have a very basic understanding of the concepts (I know what an homeomorphism is but not much more).
My idea is that a torus with one hole can be disconnected by two loops, whereas the torus with two holes may not be disconnected by two loops.
Is this argument correct? Can it me formalized (even the part on connectedness).
Or maybe there is better argument
general-topology manifolds connectedness
asked 3 hours ago
tiintin
212
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up vote
4
down vote
favorite
I would like to understand why the torus with one hole is not homeomorphic to the torus with two holes.
I have a very basic understanding of the concepts (I know what an homeomorphism is but not much more).
My idea is that a torus with one hole can be disconnected by two loops, whereas the torus with two holes may not be disconnected by two loops.
Is this argument correct? Can it me formalized (even the part on connectedness).
Or maybe there is better argument
general-topology manifolds connectedness
asked 3 hours ago
tiintin
212
New contributor
tiintin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Your idea is ok. Formally you should convince yourself that these objects have different homeomorphic invariants. For example you could consider the fundamental group. It is generated by two cycles for the torus with one hole and by more than two for the torus with two holes. Start by drawing a picture!
– James
3 hours ago
Your idea is perfect, though it needs some refinement. The torus with two holes can actually be disconnected by a single curve (can you find it?). You want to consider what are called non-separating curves — curves that don’t immediately cut the surface into separate pieces.
– Santana Afton
3 hours ago
@James I don't know what the fundamental group is, and I think it is outside the scope of the class I follow.
– tiintin
3 hours ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I would like to understand why the torus with one hole is not homeomorphic to the torus with two holes.
I have a very basic understanding of the concepts (I know what an homeomorphism is but not much more).
My idea is that a torus with one hole can be disconnected by two loops, whereas the torus with two holes may not be disconnected by two loops.
Is this argument correct? Can it me formalized (even the part on connectedness).
Or maybe there is better argument
general-topology manifolds connectedness
asked 3 hours ago
tiintin
212
New contributor
tiintin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I would like to understand why the torus with one hole is not homeomorphic to the torus with two holes.
I have a very basic understanding of the concepts (I know what an homeomorphism is but not much more).
My idea is that a torus with one hole can be disconnected by two loops, whereas the torus with two holes may not be disconnected by two loops.
Is this argument correct? Can it me formalized (even the part on connectedness).
Or maybe there is better argument
general-topology manifolds connectedness
general-topology manifolds connectedness
asked 3 hours ago
tiintin
212
New contributor
tiintin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
tiintin
212
New contributor
tiintin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
tiintin
212
New contributor
tiintin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
tiintin
212
asked 3 hours ago
tiintin
212
212
New contributor
tiintin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
tiintin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
tiintin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Your idea is ok. Formally you should convince yourself that these objects have different homeomorphic invariants. For example you could consider the fundamental group. It is generated by two cycles for the torus with one hole and by more than two for the torus with two holes. Start by drawing a picture!
– James
3 hours ago
Your idea is perfect, though it needs some refinement. The torus with two holes can actually be disconnected by a single curve (can you find it?). You want to consider what are called non-separating curves — curves that don’t immediately cut the surface into separate pieces.
– Santana Afton
3 hours ago
@James I don't know what the fundamental group is, and I think it is outside the scope of the class I follow.
– tiintin
3 hours ago
add a comment |Â
Your idea is ok. Formally you should convince yourself that these objects have different homeomorphic invariants. For example you could consider the fundamental group. It is generated by two cycles for the torus with one hole and by more than two for the torus with two holes. Start by drawing a picture!
– James
3 hours ago
Your idea is perfect, though it needs some refinement. The torus with two holes can actually be disconnected by a single curve (can you find it?). You want to consider what are called non-separating curves — curves that don’t immediately cut the surface into separate pieces.
– Santana Afton
3 hours ago
@James I don't know what the fundamental group is, and I think it is outside the scope of the class I follow.
– tiintin
3 hours ago
Your idea is ok. Formally you should convince yourself that these objects have different homeomorphic invariants. For example you could consider the fundamental group. It is generated by two cycles for the torus with one hole and by more than two for the torus with two holes. Start by drawing a picture!
– James
3 hours ago
Your idea is ok. Formally you should convince yourself that these objects have different homeomorphic invariants. For example you could consider the fundamental group. It is generated by two cycles for the torus with one hole and by more than two for the torus with two holes. Start by drawing a picture!
– James
3 hours ago
Your idea is perfect, though it needs some refinement. The torus with two holes can actually be disconnected by a single curve (can you find it?). You want to consider what are called non-separating curves — curves that don’t immediately cut the surface into separate pieces.
– Santana Afton
3 hours ago
Your idea is perfect, though it needs some refinement. The torus with two holes can actually be disconnected by a single curve (can you find it?). You want to consider what are called non-separating curves — curves that don’t immediately cut the surface into separate pieces.
– Santana Afton
3 hours ago
@James I don't know what the fundamental group is, and I think it is outside the scope of the class I follow.
– tiintin
3 hours ago
@James I don't know what the fundamental group is, and I think it is outside the scope of the class I follow.
– tiintin
3 hours ago
add a comment |Â
3 Answers
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up vote
2
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Let’s prove it!
For the sake of getting some contradiction, let $f:S_1to S_2$ be a homeomorphism. Here, $S_1$ is the one-holed torus and $S_2$ is the two-holed torus.
Now, define a non-separating curve $gamma$ on a surface $S$ to be a curve such that $Ssetminusgamma$ is still connected.
Let $gamma_1$ and $gamma_2$ be the two most popular curves on $S_1$ that cut the torus into a disk. Since $f$ is a homeomorphism, the image of these curves under $f$ (i.e. $f(gamma_1)$ and $f(gamma_2)$) must also cut $S_2$ into a disk. Now, I claim that $f(gamma_1)$ and $f(gamma_2)$ must also be non-separating in $S_2$. This is a good exercise to stew on by yourself!
Moreover, I claim that no pair of non-separating curves can cut $S_2$ into a disk. To prove this, you pick your favorite pair of non-separating curves on $S_2$, and cut along them. Show this is not a disk. Then, for any other pair of non-separating curves $alpha_i$, there is a homeomorphism taking the $alpha_i$ curves to your favorite curves. Thus, cutting along the $alpha_i$ curves cannot yield a disk.
However, from earlier we know that $f(gamma_i)$ must cut $S_2$ into a disk, thus obtaining our contradiction.
answered 2 hours ago
Santana Afton
2,1991527
Thanks you for your detailed explanations. However I have a hard time trying to figure out those popular curves. Are they the two loops that "generate" the torus? To me they cut the torus into a disc with a hole and a cylindre.
– tiintin
2 hours ago
These curves are the blue and red curves in this image. Let me know if these makes things more clear!
– Santana Afton
2 hours ago
Ok these are indeed the curves I was refering to as "generating" the torus (I guess this is not the correct vocabulary). However I still cannot see why they cut the torus into a disk...
– tiintin
2 hours ago
People do use that vocabulary to refer to those curves, but usually within the context of the fundamental group. I just wanted to be explicit. You can cut the torus in steps. What do you get if you cut along just the red curve?
– Santana Afton
2 hours ago
Oh ok I get it, you need to cut along both! (sorry, English isn't my first language).
– tiintin
2 hours ago
 |Â
show 1 more comment
up vote
1
down vote
One of these has $H^1(Sigma_1,Bbb Z)cong Bbb Z^2$ and the other $H^1(Sigma_2,Bbb Z)cong Bbb Z^4$.
In general the $g$-holed torus $Sigma_g$ has $H^1(Sigma_g,Bbb Z)cong Bbb Z^2g$.
One can see this by considering a representative loop for each of the $2g$-homotopy classes on the $g$-holed torus, and cutting along these, and by homeomorphism taking any pair of such cut surfaces to one another.
Informally you can simply fix a base-point for loops to start and end at. On the $1$-holed torus $Sigma_1$ there are two topologically distinct types of loops. Those that go 'through' the hole, and those that go around the torus. Similarly on the $2$-holed torus, there are $4$ different types of generating loops, either going 'around' each hole, or through it.
These generating loops all contribute $1$ rank to $H^1(Sigma_g,Bbb Z)$. So really there is nothing scary about this.
answered 3 hours ago
Rob
264
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Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I don't understand the notations, and I am looking for a formal proof using basic concepts. I don't know what an homotopy class is.
– tiintin
3 hours ago
The OP stated that they have a very basic understanding of surface topology. I doubt that an appeal to homology would be of much use, here.
– Santana Afton
3 hours ago
add a comment |Â
up vote
0
down vote
You can get a figure $8$ as the intersection of a plane with the torus with two holes but not with the one with only one hole.
answered 3 hours ago
Mohammad Riazi-Kermani
34.9k41855
Ok I get the idea, could you please develop it formally?
– tiintin
3 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Let’s prove it!
For the sake of getting some contradiction, let $f:S_1to S_2$ be a homeomorphism. Here, $S_1$ is the one-holed torus and $S_2$ is the two-holed torus.
Now, define a non-separating curve $gamma$ on a surface $S$ to be a curve such that $Ssetminusgamma$ is still connected.
Let $gamma_1$ and $gamma_2$ be the two most popular curves on $S_1$ that cut the torus into a disk. Since $f$ is a homeomorphism, the image of these curves under $f$ (i.e. $f(gamma_1)$ and $f(gamma_2)$) must also cut $S_2$ into a disk. Now, I claim that $f(gamma_1)$ and $f(gamma_2)$ must also be non-separating in $S_2$. This is a good exercise to stew on by yourself!
Moreover, I claim that no pair of non-separating curves can cut $S_2$ into a disk. To prove this, you pick your favorite pair of non-separating curves on $S_2$, and cut along them. Show this is not a disk. Then, for any other pair of non-separating curves $alpha_i$, there is a homeomorphism taking the $alpha_i$ curves to your favorite curves. Thus, cutting along the $alpha_i$ curves cannot yield a disk.
However, from earlier we know that $f(gamma_i)$ must cut $S_2$ into a disk, thus obtaining our contradiction.
answered 2 hours ago
Santana Afton
2,1991527
Thanks you for your detailed explanations. However I have a hard time trying to figure out those popular curves. Are they the two loops that "generate" the torus? To me they cut the torus into a disc with a hole and a cylindre.
– tiintin
2 hours ago
These curves are the blue and red curves in this image. Let me know if these makes things more clear!
– Santana Afton
2 hours ago
Ok these are indeed the curves I was refering to as "generating" the torus (I guess this is not the correct vocabulary). However I still cannot see why they cut the torus into a disk...
– tiintin
2 hours ago
People do use that vocabulary to refer to those curves, but usually within the context of the fundamental group. I just wanted to be explicit. You can cut the torus in steps. What do you get if you cut along just the red curve?
– Santana Afton
2 hours ago
Oh ok I get it, you need to cut along both! (sorry, English isn't my first language).
– tiintin
2 hours ago
 |Â
show 1 more comment
up vote
2
down vote
Let’s prove it!
For the sake of getting some contradiction, let $f:S_1to S_2$ be a homeomorphism. Here, $S_1$ is the one-holed torus and $S_2$ is the two-holed torus.
Now, define a non-separating curve $gamma$ on a surface $S$ to be a curve such that $Ssetminusgamma$ is still connected.
Let $gamma_1$ and $gamma_2$ be the two most popular curves on $S_1$ that cut the torus into a disk. Since $f$ is a homeomorphism, the image of these curves under $f$ (i.e. $f(gamma_1)$ and $f(gamma_2)$) must also cut $S_2$ into a disk. Now, I claim that $f(gamma_1)$ and $f(gamma_2)$ must also be non-separating in $S_2$. This is a good exercise to stew on by yourself!
Moreover, I claim that no pair of non-separating curves can cut $S_2$ into a disk. To prove this, you pick your favorite pair of non-separating curves on $S_2$, and cut along them. Show this is not a disk. Then, for any other pair of non-separating curves $alpha_i$, there is a homeomorphism taking the $alpha_i$ curves to your favorite curves. Thus, cutting along the $alpha_i$ curves cannot yield a disk.
However, from earlier we know that $f(gamma_i)$ must cut $S_2$ into a disk, thus obtaining our contradiction.
answered 2 hours ago
Santana Afton
2,1991527
Thanks you for your detailed explanations. However I have a hard time trying to figure out those popular curves. Are they the two loops that "generate" the torus? To me they cut the torus into a disc with a hole and a cylindre.
– tiintin
2 hours ago
These curves are the blue and red curves in this image. Let me know if these makes things more clear!
– Santana Afton
2 hours ago
Ok these are indeed the curves I was refering to as "generating" the torus (I guess this is not the correct vocabulary). However I still cannot see why they cut the torus into a disk...
– tiintin
2 hours ago
People do use that vocabulary to refer to those curves, but usually within the context of the fundamental group. I just wanted to be explicit. You can cut the torus in steps. What do you get if you cut along just the red curve?
– Santana Afton
2 hours ago
Oh ok I get it, you need to cut along both! (sorry, English isn't my first language).
– tiintin
2 hours ago
 |Â
show 1 more comment
up vote
2
down vote
up vote
2
down vote
Let’s prove it!
For the sake of getting some contradiction, let $f:S_1to S_2$ be a homeomorphism. Here, $S_1$ is the one-holed torus and $S_2$ is the two-holed torus.
Now, define a non-separating curve $gamma$ on a surface $S$ to be a curve such that $Ssetminusgamma$ is still connected.
Let $gamma_1$ and $gamma_2$ be the two most popular curves on $S_1$ that cut the torus into a disk. Since $f$ is a homeomorphism, the image of these curves under $f$ (i.e. $f(gamma_1)$ and $f(gamma_2)$) must also cut $S_2$ into a disk. Now, I claim that $f(gamma_1)$ and $f(gamma_2)$ must also be non-separating in $S_2$. This is a good exercise to stew on by yourself!
Moreover, I claim that no pair of non-separating curves can cut $S_2$ into a disk. To prove this, you pick your favorite pair of non-separating curves on $S_2$, and cut along them. Show this is not a disk. Then, for any other pair of non-separating curves $alpha_i$, there is a homeomorphism taking the $alpha_i$ curves to your favorite curves. Thus, cutting along the $alpha_i$ curves cannot yield a disk.
However, from earlier we know that $f(gamma_i)$ must cut $S_2$ into a disk, thus obtaining our contradiction.
answered 2 hours ago
Santana Afton
2,1991527
Let’s prove it!
For the sake of getting some contradiction, let $f:S_1to S_2$ be a homeomorphism. Here, $S_1$ is the one-holed torus and $S_2$ is the two-holed torus.
Now, define a non-separating curve $gamma$ on a surface $S$ to be a curve such that $Ssetminusgamma$ is still connected.
Let $gamma_1$ and $gamma_2$ be the two most popular curves on $S_1$ that cut the torus into a disk. Since $f$ is a homeomorphism, the image of these curves under $f$ (i.e. $f(gamma_1)$ and $f(gamma_2)$) must also cut $S_2$ into a disk. Now, I claim that $f(gamma_1)$ and $f(gamma_2)$ must also be non-separating in $S_2$. This is a good exercise to stew on by yourself!
Moreover, I claim that no pair of non-separating curves can cut $S_2$ into a disk. To prove this, you pick your favorite pair of non-separating curves on $S_2$, and cut along them. Show this is not a disk. Then, for any other pair of non-separating curves $alpha_i$, there is a homeomorphism taking the $alpha_i$ curves to your favorite curves. Thus, cutting along the $alpha_i$ curves cannot yield a disk.
However, from earlier we know that $f(gamma_i)$ must cut $S_2$ into a disk, thus obtaining our contradiction.
answered 2 hours ago
Santana Afton
2,1991527
edited 2 hours ago
answered 2 hours ago
Santana Afton
2,1991527
answered 2 hours ago
Santana Afton
2,1991527
answered 2 hours ago
Santana Afton
2,1991527
2,1991527
Thanks you for your detailed explanations. However I have a hard time trying to figure out those popular curves. Are they the two loops that "generate" the torus? To me they cut the torus into a disc with a hole and a cylindre.
– tiintin
2 hours ago
These curves are the blue and red curves in this image. Let me know if these makes things more clear!
– Santana Afton
2 hours ago
Ok these are indeed the curves I was refering to as "generating" the torus (I guess this is not the correct vocabulary). However I still cannot see why they cut the torus into a disk...
– tiintin
2 hours ago
People do use that vocabulary to refer to those curves, but usually within the context of the fundamental group. I just wanted to be explicit. You can cut the torus in steps. What do you get if you cut along just the red curve?
– Santana Afton
2 hours ago
Oh ok I get it, you need to cut along both! (sorry, English isn't my first language).
– tiintin
2 hours ago
 |Â
show 1 more comment
Thanks you for your detailed explanations. However I have a hard time trying to figure out those popular curves. Are they the two loops that "generate" the torus? To me they cut the torus into a disc with a hole and a cylindre.
– tiintin
2 hours ago
These curves are the blue and red curves in this image. Let me know if these makes things more clear!
– Santana Afton
2 hours ago
Ok these are indeed the curves I was refering to as "generating" the torus (I guess this is not the correct vocabulary). However I still cannot see why they cut the torus into a disk...
– tiintin
2 hours ago
People do use that vocabulary to refer to those curves, but usually within the context of the fundamental group. I just wanted to be explicit. You can cut the torus in steps. What do you get if you cut along just the red curve?
– Santana Afton
2 hours ago
Oh ok I get it, you need to cut along both! (sorry, English isn't my first language).
– tiintin
2 hours ago
Thanks you for your detailed explanations. However I have a hard time trying to figure out those popular curves. Are they the two loops that "generate" the torus? To me they cut the torus into a disc with a hole and a cylindre.
– tiintin
2 hours ago
Thanks you for your detailed explanations. However I have a hard time trying to figure out those popular curves. Are they the two loops that "generate" the torus? To me they cut the torus into a disc with a hole and a cylindre.
– tiintin
2 hours ago
These curves are the blue and red curves in this image. Let me know if these makes things more clear!
– Santana Afton
2 hours ago
These curves are the blue and red curves in this image. Let me know if these makes things more clear!
– Santana Afton
2 hours ago
Ok these are indeed the curves I was refering to as "generating" the torus (I guess this is not the correct vocabulary). However I still cannot see why they cut the torus into a disk...
– tiintin
2 hours ago
Ok these are indeed the curves I was refering to as "generating" the torus (I guess this is not the correct vocabulary). However I still cannot see why they cut the torus into a disk...
– tiintin
2 hours ago
People do use that vocabulary to refer to those curves, but usually within the context of the fundamental group. I just wanted to be explicit. You can cut the torus in steps. What do you get if you cut along just the red curve?
– Santana Afton
2 hours ago
People do use that vocabulary to refer to those curves, but usually within the context of the fundamental group. I just wanted to be explicit. You can cut the torus in steps. What do you get if you cut along just the red curve?
– Santana Afton
2 hours ago
Oh ok I get it, you need to cut along both! (sorry, English isn't my first language).
– tiintin
2 hours ago
Oh ok I get it, you need to cut along both! (sorry, English isn't my first language).
– tiintin
2 hours ago
 |Â
show 1 more comment
up vote
1
down vote
One of these has $H^1(Sigma_1,Bbb Z)cong Bbb Z^2$ and the other $H^1(Sigma_2,Bbb Z)cong Bbb Z^4$.
In general the $g$-holed torus $Sigma_g$ has $H^1(Sigma_g,Bbb Z)cong Bbb Z^2g$.
One can see this by considering a representative loop for each of the $2g$-homotopy classes on the $g$-holed torus, and cutting along these, and by homeomorphism taking any pair of such cut surfaces to one another.
Informally you can simply fix a base-point for loops to start and end at. On the $1$-holed torus $Sigma_1$ there are two topologically distinct types of loops. Those that go 'through' the hole, and those that go around the torus. Similarly on the $2$-holed torus, there are $4$ different types of generating loops, either going 'around' each hole, or through it.
These generating loops all contribute $1$ rank to $H^1(Sigma_g,Bbb Z)$. So really there is nothing scary about this.
answered 3 hours ago
Rob
264
New contributor
Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I don't understand the notations, and I am looking for a formal proof using basic concepts. I don't know what an homotopy class is.
– tiintin
3 hours ago
The OP stated that they have a very basic understanding of surface topology. I doubt that an appeal to homology would be of much use, here.
– Santana Afton
3 hours ago
add a comment |Â
up vote
1
down vote
One of these has $H^1(Sigma_1,Bbb Z)cong Bbb Z^2$ and the other $H^1(Sigma_2,Bbb Z)cong Bbb Z^4$.
In general the $g$-holed torus $Sigma_g$ has $H^1(Sigma_g,Bbb Z)cong Bbb Z^2g$.
One can see this by considering a representative loop for each of the $2g$-homotopy classes on the $g$-holed torus, and cutting along these, and by homeomorphism taking any pair of such cut surfaces to one another.
Informally you can simply fix a base-point for loops to start and end at. On the $1$-holed torus $Sigma_1$ there are two topologically distinct types of loops. Those that go 'through' the hole, and those that go around the torus. Similarly on the $2$-holed torus, there are $4$ different types of generating loops, either going 'around' each hole, or through it.
These generating loops all contribute $1$ rank to $H^1(Sigma_g,Bbb Z)$. So really there is nothing scary about this.
answered 3 hours ago
Rob
264
New contributor
Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I don't understand the notations, and I am looking for a formal proof using basic concepts. I don't know what an homotopy class is.
– tiintin
3 hours ago
The OP stated that they have a very basic understanding of surface topology. I doubt that an appeal to homology would be of much use, here.
– Santana Afton
3 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
One of these has $H^1(Sigma_1,Bbb Z)cong Bbb Z^2$ and the other $H^1(Sigma_2,Bbb Z)cong Bbb Z^4$.
In general the $g$-holed torus $Sigma_g$ has $H^1(Sigma_g,Bbb Z)cong Bbb Z^2g$.
One can see this by considering a representative loop for each of the $2g$-homotopy classes on the $g$-holed torus, and cutting along these, and by homeomorphism taking any pair of such cut surfaces to one another.
Informally you can simply fix a base-point for loops to start and end at. On the $1$-holed torus $Sigma_1$ there are two topologically distinct types of loops. Those that go 'through' the hole, and those that go around the torus. Similarly on the $2$-holed torus, there are $4$ different types of generating loops, either going 'around' each hole, or through it.
These generating loops all contribute $1$ rank to $H^1(Sigma_g,Bbb Z)$. So really there is nothing scary about this.
answered 3 hours ago
Rob
264
New contributor
Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
One of these has $H^1(Sigma_1,Bbb Z)cong Bbb Z^2$ and the other $H^1(Sigma_2,Bbb Z)cong Bbb Z^4$.
In general the $g$-holed torus $Sigma_g$ has $H^1(Sigma_g,Bbb Z)cong Bbb Z^2g$.
One can see this by considering a representative loop for each of the $2g$-homotopy classes on the $g$-holed torus, and cutting along these, and by homeomorphism taking any pair of such cut surfaces to one another.
Informally you can simply fix a base-point for loops to start and end at. On the $1$-holed torus $Sigma_1$ there are two topologically distinct types of loops. Those that go 'through' the hole, and those that go around the torus. Similarly on the $2$-holed torus, there are $4$ different types of generating loops, either going 'around' each hole, or through it.
These generating loops all contribute $1$ rank to $H^1(Sigma_g,Bbb Z)$. So really there is nothing scary about this.
answered 3 hours ago
Rob
264
New contributor
Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
answered 3 hours ago
Rob
264
New contributor
Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
Rob
264
answered 3 hours ago
Rob
264
264
New contributor
Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Rob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I don't understand the notations, and I am looking for a formal proof using basic concepts. I don't know what an homotopy class is.
– tiintin
3 hours ago
The OP stated that they have a very basic understanding of surface topology. I doubt that an appeal to homology would be of much use, here.
– Santana Afton
3 hours ago
add a comment |Â
I don't understand the notations, and I am looking for a formal proof using basic concepts. I don't know what an homotopy class is.
– tiintin
3 hours ago
The OP stated that they have a very basic understanding of surface topology. I doubt that an appeal to homology would be of much use, here.
– Santana Afton
3 hours ago
I don't understand the notations, and I am looking for a formal proof using basic concepts. I don't know what an homotopy class is.
– tiintin
3 hours ago
I don't understand the notations, and I am looking for a formal proof using basic concepts. I don't know what an homotopy class is.
– tiintin
3 hours ago
The OP stated that they have a very basic understanding of surface topology. I doubt that an appeal to homology would be of much use, here.
– Santana Afton
3 hours ago
The OP stated that they have a very basic understanding of surface topology. I doubt that an appeal to homology would be of much use, here.
– Santana Afton
3 hours ago
add a comment |Â
up vote
0
down vote
You can get a figure $8$ as the intersection of a plane with the torus with two holes but not with the one with only one hole.
answered 3 hours ago
Mohammad Riazi-Kermani
34.9k41855
Ok I get the idea, could you please develop it formally?
– tiintin
3 hours ago
add a comment |Â
up vote
0
down vote
You can get a figure $8$ as the intersection of a plane with the torus with two holes but not with the one with only one hole.
answered 3 hours ago
Mohammad Riazi-Kermani
34.9k41855
Ok I get the idea, could you please develop it formally?
– tiintin
3 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can get a figure $8$ as the intersection of a plane with the torus with two holes but not with the one with only one hole.
answered 3 hours ago
Mohammad Riazi-Kermani
34.9k41855
You can get a figure $8$ as the intersection of a plane with the torus with two holes but not with the one with only one hole.
answered 3 hours ago
Mohammad Riazi-Kermani
34.9k41855
answered 3 hours ago
Mohammad Riazi-Kermani
34.9k41855
answered 3 hours ago
Mohammad Riazi-Kermani
34.9k41855
answered 3 hours ago
Mohammad Riazi-Kermani
34.9k41855
34.9k41855
Ok I get the idea, could you please develop it formally?
– tiintin
3 hours ago
add a comment |Â
Ok I get the idea, could you please develop it formally?
– tiintin
3 hours ago
Ok I get the idea, could you please develop it formally?
– tiintin
3 hours ago
Ok I get the idea, could you please develop it formally?
– tiintin
3 hours ago
add a comment |Â
tiintin is a new contributor. Be nice, and check out our Code of Conduct.
tiintin is a new contributor. Be nice, and check out our Code of Conduct.
tiintin is a new contributor. Be nice, and check out our Code of Conduct.
tiintin is a new contributor. Be nice, and check out our Code of Conduct.
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Your idea is ok. Formally you should convince yourself that these objects have different homeomorphic invariants. For example you could consider the fundamental group. It is generated by two cycles for the torus with one hole and by more than two for the torus with two holes. Start by drawing a picture!
– James
3 hours ago
Your idea is perfect, though it needs some refinement. The torus with two holes can actually be disconnected by a single curve (can you find it?). You want to consider what are called non-separating curves — curves that don’t immediately cut the surface into separate pieces.
– Santana Afton
3 hours ago
@James I don't know what the fundamental group is, and I think it is outside the scope of the class I follow.
– tiintin
3 hours ago